LeetCode #3835 — MEDIUM

Count Subarrays With Cost Less Than or Equal to K

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums, and an integer k.

For any subarray nums[l..r], define its cost as:

cost = (max(nums[l..r]) - min(nums[l..r])) * (r - l + 1).

Return an integer denoting the number of subarrays of nums whose cost is less than or equal to k.

Example 1:

Input: nums = [1,3,2], k = 4

Output: 5

Explanation:

We consider all subarrays of nums:

nums[0..0]: cost = (1 - 1) * 1 = 0
nums[0..1]: cost = (3 - 1) * 2 = 4
nums[0..2]: cost = (3 - 1) * 3 = 6
nums[1..1]: cost = (3 - 3) * 1 = 0
nums[1..2]: cost = (3 - 2) * 2 = 2
nums[2..2]: cost = (2 - 2) * 1 = 0

There are 5 subarrays whose cost is less than or equal to 4.

Example 2:

Input: nums = [5,5,5,5], k = 0

Output: 10

Explanation:

For any subarray of nums, the maximum and minimum values are the same, so the cost is always 0.

As a result, every subarray of nums has cost less than or equal to 0.

For an array of length 4, the total number of subarrays is (4 * 5) / 2 = 10.

Example 3:

Input: nums = [1,2,3], k = 0

Output: 3

Explanation:

The only subarrays of nums with cost 0 are the single-element subarrays, and there are 3 of them.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
0 <= k <= 10¹⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums, and an integer k. For any subarray nums[l..r], define its cost as: cost = (max(nums[l..r]) - min(nums[l..r])) * (r - l + 1). Return an integer denoting the number of subarrays of nums whose cost is less than or equal to k.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

[1,3,2]
4

Example 2

[5,5,5,5]
0

Example 3

[1,2,3]
0

Step 02

Core Insight

What unlocks the optimal approach

Use sliding window.
A monotonic deque is useful for maintaining the maximum or minimum in <code>O(1)</code> time per operation.
When the current window's cost exceeds <code>k</code>, move the left boundary forward until the window becomes valid again. Count all subarrays ending at the current right boundary.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3835: Count Subarrays With Cost Less Than or Equal to K
class Solution {
    public long countSubarrays(int[] nums, long k) {
        long ans = 0;
        Deque<Integer> q1 = new ArrayDeque<>();
        Deque<Integer> q2 = new ArrayDeque<>();
        int l = 0;

        for (int r = 0; r < nums.length; r++) {
            int x = nums[r];

            while (!q1.isEmpty() && nums[q1.peekLast()] <= x) {
                q1.pollLast();
            }
            while (!q2.isEmpty() && nums[q2.peekLast()] >= x) {
                q2.pollLast();
            }
            q1.addLast(r);
            q2.addLast(r);

            while (
                l < r && (long) (nums[q1.peekFirst()] - nums[q2.peekFirst()]) * (r - l + 1) > k) {
                l++;
                if (q1.peekFirst() < l) {
                    q1.pollFirst();
                }
                if (q2.peekFirst() < l) {
                    q2.pollFirst();
                }
            }

            ans += r - l + 1;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3835: Count Subarrays With Cost Less Than or Equal to K
func countSubarrays(nums []int, k int64) int64 {
	var ans int64 = 0
	q1 := make([]int, 0)
	q2 := make([]int, 0)
	l := 0

	for r, x := range nums {
		for len(q1) > 0 && nums[q1[len(q1)-1]] <= x {
			q1 = q1[:len(q1)-1]
		}
		for len(q2) > 0 && nums[q2[len(q2)-1]] >= x {
			q2 = q2[:len(q2)-1]
		}
		q1 = append(q1, r)
		q2 = append(q2, r)

		for l < r &&
			int64(nums[q1[0]]-nums[q2[0]])*int64(r-l+1) > k {
			l++
			if q1[0] < l {
				q1 = q1[1:]
			}
			if q2[0] < l {
				q2 = q2[1:]
			}
		}
		ans += int64(r - l + 1)
	}
	return ans
}

# Accepted solution for LeetCode #3835: Count Subarrays With Cost Less Than or Equal to K
class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        ans = 0
        q1 = deque()
        q2 = deque()
        l = 0
        for r, x in enumerate(nums):
            while q1 and nums[q1[-1]] <= x:
                q1.pop()
            while q2 and nums[q2[-1]] >= x:
                q2.pop()
            q1.append(r)
            q2.append(r)
            while l < r and (nums[q1[0]] - nums[q2[0]]) * (r - l + 1) > k:
                l += 1
                if q1[0] < l:
                    q1.popleft()
                if q2[0] < l:
                    q2.popleft()
            ans += r - l + 1
        return ans

// Accepted solution for LeetCode #3835: Count Subarrays With Cost Less Than or Equal to K
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3835: Count Subarrays With Cost Less Than or Equal to K
// class Solution {
//     public long countSubarrays(int[] nums, long k) {
//         long ans = 0;
//         Deque<Integer> q1 = new ArrayDeque<>();
//         Deque<Integer> q2 = new ArrayDeque<>();
//         int l = 0;
// 
//         for (int r = 0; r < nums.length; r++) {
//             int x = nums[r];
// 
//             while (!q1.isEmpty() && nums[q1.peekLast()] <= x) {
//                 q1.pollLast();
//             }
//             while (!q2.isEmpty() && nums[q2.peekLast()] >= x) {
//                 q2.pollLast();
//             }
//             q1.addLast(r);
//             q2.addLast(r);
// 
//             while (
//                 l < r && (long) (nums[q1.peekFirst()] - nums[q2.peekFirst()]) * (r - l + 1) > k) {
//                 l++;
//                 if (q1.peekFirst() < l) {
//                     q1.pollFirst();
//                 }
//                 if (q2.peekFirst() < l) {
//                     q2.pollFirst();
//                 }
//             }
// 
//             ans += r - l + 1;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3835: Count Subarrays With Cost Less Than or Equal to K
function countSubarrays(nums: number[], k: number): number {
    let ans = 0;
    const q1: number[] = [];
    const q2: number[] = [];
    let h1 = 0,
        t1 = 0;
    let h2 = 0,
        t2 = 0;
    let l = 0;
    for (let r = 0; r < nums.length; r++) {
        const x = nums[r];
        while (h1 < t1 && nums[q1[t1 - 1]] <= x) {
            t1--;
        }
        while (h2 < t2 && nums[q2[t2 - 1]] >= x) {
            t2--;
        }
        q1[t1++] = r;
        q2[t2++] = r;
        while (l < r && (nums[q1[h1]] - nums[q2[h2]]) * (r - l + 1) > k) {
            l++;
            if (q1[h1] < l) {
                h1++;
            }
            if (q2[h2] < l) {
                h2++;
            }
        }
        ans += r - l + 1;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.